Calculating Emptying Pipe Time
This problem involves calculating the time taken by an emptying pipe based on the filling times of two inlet pipes and the combined filling time when all three pipes are open.
Pipe Rates Calculation
- Let the time taken by the first pipe to fill the cistern be $T_1 = 36$ minutes. Its filling rate is $R_1 = \frac{1}{T_1} = \frac{1}{36}$ cistern per minute.
- Let the time taken by the second pipe to fill the cistern be $T_2 = 45$ minutes. Its filling rate is $R_2 = \frac{1}{T_2} = \frac{1}{45}$ cistern per minute.
- Let the time taken by the bottom pipe to empty the cistern be $T_3$ minutes. Its emptying rate is $R_3 = \frac{1}{T_3}$ cistern per minute. Since it empties, we consider this rate as negative in the combined calculation.
- When all three pipes are open, the cistern fills in $T_{net} = 70$ minutes. The net filling rate is $R_{net} = \frac{1}{70}$ cistern per minute.
Combined Rate Equation
The net rate is the sum of the filling rates minus the emptying rate:
$R_{net} = R_1 + R_2 - R_3$
Substituting the known values:
$\frac{1}{70} = \frac{1}{36} + \frac{1}{45} - R_3$
Solving for the Emptying Rate ($R_3$)
First, find the combined rate of the two filling pipes:
$R_1 + R_2 = \frac{1}{36} + \frac{1}{45}$
The least common multiple (LCM) of 36 and 45 is 180.
$R_1 + R_2 = \frac{5}{180} + \frac{4}{180} = \frac{9}{180} = \frac{1}{20}$ cistern per minute.
Now substitute this back into the net rate equation:
$\frac{1}{70} = \frac{1}{20} - R_3$
Rearrange to solve for $R_3$:
$R_3 = \frac{1}{20} - \frac{1}{70}$
The LCM of 20 and 70 is 140.
$R_3 = \frac{7}{140} - \frac{2}{140} = \frac{5}{140} = \frac{1}{28}$ cistern per minute.
Final Answer: Time to Empty
The time taken by the bottom pipe to empty the completely filled cistern is the reciprocal of its rate $R_3$:
$T_3 = \frac{1}{R_3} = \frac{1}{1/28} = 28$ minutes.
Similar Questions
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Important Questions from Pipe and Cistern
A pipe can fill a cistern in 20 minutes where as the cistern when full can be emptied by a leak in 28 minutes. When both are opened, The time taken to fill the cistern is:
‘A’ pipe can empty a tank in 20 minutes. The second pipe ‘B’ has a diameter twice as that of ‘A’. If both A & B pipe are attached to the tank how much time will be required to empty the tank?
Pipes A and B can empty a full tank in 16 hours and 24 hours, respectively. Pipe C alone can fill the empty tank in 4 hours. If A, B and C are opened together, the tank will be 35% full after :
Pipes A and B can fill a tank in 36 minutes and 45 minutes, respectively. Both these pipes were opened simultaneously. After 20 minutes, a leak at the bottom of the tank was spotted which was immediately sealed. The tank was full in another 15 minutes. The leak alone can empty the full tank in:
A cistern has a leak which would empty it in 6 hours. A tap is turned on which admits 10 litres of water per minute into the cistern. When it is full it is now emptied in 10 hours. What is the capacity (in litres) of the cistern?
